Answer
$$
\cos (x^{2}-x)=x^{4}
$$
The roots of the given equation, correct to eight decimal places, are -0.73485910 and 1.
Work Step by Step
$$
\cos (x^{2}-x)=x^{4}
$$
We first rewrite the equation in standard form:
$$
\cos (x^{2}-x)-x^{4}=0
$$
Therefore we let
$$
f(x)=\cos (x^{2}-x)-x^{4}=0
$$
Then
$$
f^{'}(x)=-(2x-1)\sin(x^{2}-x)-4x^{3}
$$
so, Formula 2 (Newton’s method) becomes
$$
\begin{aligned} x_{n+1} &=x_{n}-\frac{f(x)}{f^{'}(x)}\\
\\ &=x_{n}-\frac{ \cos (x_{n}^{2}-x_{n})-x_{n}^{4} }{-(2x_{n}-1)\sin(x_{n}^{2}-x_{n})-4x_{n}^{3}}\\
&=x_{n}+\frac{ \cos (x_{n}^{2}-x_{n})-x_{n}^{4} }{(2x_{n}-1)\sin(x_{n}^{2}-x_{n})+4x_{n}^{3}}
\end{aligned}
$$
In order to guess a suitable value for $x_1$ we sketch the graphs of $y=\cos (x^{2}-x) $ and $y = x^{4} $ in the Figure .
It appears that they intersect at a point whose x-coordinate is somewhat less than -0.7, so let’s take $x_{1} =- 0.7 $ as a convenient first approximation. Then Newton’s method gives
$$
\begin{aligned} x_{2} &=x_{1}+\frac{ \cos (x_{1}^{2}-x_{1})-x_{1}^{4} }{(2x_{1}-1)\sin(x_{1}^{2}-x_{1})+4x_{1}^{3}} \\
&=(-0.7)+\frac{ \cos ((-0.7)^{2}-(-0.7))-(-0.7)^{4} }{(2(-0.7)-1)\sin((-0.7)^{2}-(-0.7))+4(-0.7)^{3}} \approx -0.73654354
\end{aligned}
$$
repeating we get
$$
x_{3}\approx -0.73486274
x_{4} \approx -0.73485910 \approx x_{5} ,
$$
we conclude that one root of the given equation, correct to eight decimal places, is -0.73485910, and the other is exactly 1.
